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anonymous
 4 years ago
Hello everyone. It's been awhile since I've been up here. I have one about a polynomial / another polynomial. Here it is:
(x^2y^2+2yzz^2)/(x+yz)
Any thoughts?
anonymous
 4 years ago
Hello everyone. It's been awhile since I've been up here. I have one about a polynomial / another polynomial. Here it is: (x^2y^2+2yzz^2)/(x+yz) Any thoughts?

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whpalmer4
 4 years ago
Best ResponseYou've already chosen the best response.1You do long division on polynomials just like on numbers. What is the biggest thing you could multiply (x+yz) by and have the result be < or = to the numerator? After you subtract the result from the numerator, what is left? Is there another value you could multiply (x+yz) by that would be less than or equal to that? If so, that's another term in your answer, otherwise, what's left is the remainder, if it isn't 0.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's a long division problem. The book answer is xy+z. But how do you get there doing the long division?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It would be pretty easy if it was just a poly / a monomial. But this thing was tricky to me.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Also I don't see how you could subtract the give denominator from the given numerator when there aren't any like terms on one versus the other.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0A little easier said that done huh?

whpalmer4
 4 years ago
Best ResponseYou've already chosen the best response.1You gotta have faith :) \[(x^2  y^2 + 2yz  z^2) / (x + y  z)\] Okay, by looking it, we know our answer has to have an x, a y, and a z, right? That's the only way we'll get x^2, y^2 and z^2 terms. We'll do this as a "speculative division" — we'll speculate about the answer, and then see if works. What we do will also bear some similarity to solving multiple equations with combination. Let's make a little table of what {x, y, z} * (x+yz) gives us: \[x(x+yz) = x^2 + xy  xz\]\[y(x+yz) = y^2 + xy  yz\]\[z(x+yz) = z^2 + xz + yz\] Those are our building blocks. How do we combine them to get our original numerator? Well, we need an x in the quotient to produce the x^2 term. We need a y in the quotient to produce the y^2 term. We need a z in the quotient to produce the z^2 term. Let's just add up those equations and see if we get the right thing! \[x(x+yz)  y(x+yz) + z(x+yz)=\]\[x^2+xyxz  (y^2+xyyz) + (z^2+xz+yz)=\] \[x^2 + xy  xz y^2 xy (yz) z^2 +xz +yz=\]\[x^2y^2 z^2 + xy xy +xz xz +yz+yz \] and after the dust settles and we cross off all the terms that combine to 0, we have\[x^2y^2+2yz z^2\] which is our original numerator. Therefore, the quotient we "guessed", (x  y + z), is correct!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It took me a few minutes. Let me look at what you have there. Thanks for the help too.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I will have to test that theory on other problems, but thanks for that. I really appreciate it. However, do you know how to do it using the long division?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In other words, suppose I don't have a guess. Then what?

whpalmer4
 4 years ago
Best ResponseYou've already chosen the best response.1Well, it really *is* what you do when you do long division, I just organized the work a little differently.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I will try that over the weekend on some other problems.

whpalmer4
 4 years ago
Best ResponseYou've already chosen the best response.1x  y + z x+yz x^2  y^2 + 2yz  z^2 x^2 xy xz  y^2 + 2yz  z^2  xy +xz y^2 +yz  xy  yz  z^2 +xz yz z^2 +xz 

whpalmer4
 4 years ago
Best ResponseYou've already chosen the best response.1I tend to make mistakes when doing lots of algebra, so if I can think a little first and simplify the pencilpushing, my accuracy goes up :)

whpalmer4
 4 years ago
Best ResponseYou've already chosen the best response.1Writing it out like a long division problem, it took me a couple of tries before I avoided making any mistakes. Doing it the "combination" way, I did it the first time, and it took less time, too!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That was the first time I've seen it done that way. I am going to look that up on youtube later.

whpalmer4
 4 years ago
Best ResponseYou've already chosen the best response.1I agree it's pretty weird seeing all of these "other" terms popping up as you do the long division, but like I said, you just have to have faith that they will all go away in the end :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It kind of threw me for a loop, but I will have this figured out by tomorrow.

whpalmer4
 4 years ago
Best ResponseYou've already chosen the best response.1That's the attitude I like to hear! :) Too many people just say "oh, this is hard" or "I'm dumb" or whatever. Much of this stuff isn't really hard at all — it just takes the right explanation, and practice.
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